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392 Chapter 18 Free Energy and Thermodynamics Solution: T = 273.15 + (-89.5°C) = 183.65 K; = fus = 5.37 kJ/mol; 5.37 5.37 then = T = 5.37 X = for 1.00 mol 183.65 K Check: The units (J/K) are correct. The magnitude of the answer (29 J/K) makes sense because the number of kJ is much smaller than the temperature and then a factor of 10³ was applied. 18.33 Given: 45.0 g C₃H₆O, fus = 5.69 kJ/mol, T = -94.8°C Find: Conceptual plan: °C K then fus then kJ/mol J/mol then J/mol J/g then T AS 1000J 1 mol 1kJ Solution: T = 273.15 + (-94.8°C) = 178.35 K; = fus = 5.69 kJ/mol; 5.69 mol X 1 kJ = 5.69 10³ J/mol then 45.0 g X 5.69 X 10³ 1 X 58.08 1 = 4.40857 10³ J then = T = 4.40857 X 10³ J/mol = 24.7 K J for 45.0 g. Check: The units (J/K) are correct. The magnitude of the answer (25 J/K) makes sense because we have a little less than a mole and the number of kJ is much smaller than the temperature and then a factor of 10³ was applied. 18.35 (a) > 0 because a gas is being generated. (b) S 0 because 6 moles of gas are being converted to 7 moles of gas. Because AH 0 and the reaction is nonspontaneous at all temperatures. (c) 0 and the reaction is nonspontaneous at all temperatures. (d) > 0 because 9 moles of gas are being converted to 10 moles of gas. Because AH the reaction is spontaneous at all temperatures. 18.39 (a) Given: = 385 kJ, T = 298 K Find: Conceptual Plan: kJ J then T Solution: -385 kJ = -385,000J then = T = = = 1.29 X J/K Check: The units (J/K) are correct. The magnitude of the answer (10³ J/KJ) makes sense because the kJ and the temperature started with very similar values and then a factor of 10³ was applied. (b) Given: = -385 kJ, = 77 K Find: Conceptual Plan: kJ J then Solution: -385 kJ X 1 = 385,000 then = T = -(-385,000J) 77 K = 5.00 J/K Check: The units (J/K) are correct. The magnitude of the answer (5 X 10³J/K) makes sense because the tem- perature is much lower than in part (a); so the answer should increase. Copyright © 2017 Pearson Education, Inc.